Question

A newly discovered planet has a mean radius of 2020 km. A vehicle on the planet\'s surface is moving in the same direction as the planet\'s rotation, and its speedometer reads 145 km/h. If the angular velocity of the vehicle about the planet\'s center is 5.28 times as large as the angular velocity of the planet, what is the period of the planet\'s rotation?

Answer 1

@perl

Answer 2

hey

Answer 3

can you help me?

Answer 4

yes

Answer 5

lets call the period of the planet to make a complete revolution T

Answer 6

lets write some stuff down

Answer 7

the angular velocity of planet
2pi / Tp
angular velocity of vehicle
2pi / Tv

tangential velocity = radius * angular velocity

Answer 8

From v = omega x r you can get omega (angular velocity for the vehicle.
Then, you can calculate omega for the planet because omega vehicle/omega planet = 5.28
Once you have omega for the planet, omega = 2 x pi x frequency
Solve for the frequency and then
f = 1/T where T is the period

Answer 9

Tv stands for period of vehicle
Tp stands for period of planet

so we are given

2 Pi / (Tv) = 5 * 2 Pi / (Tp)

Answer 10

and tangential velocity is 145 km /hr

Answer 11

so we can find Tv

145 = (2020 ) * 2Pi / Tv

once we find Tv, we can substitute to find Tp

Answer 12

Tv = 87.53 hours is the period of the vehicle

Answer 13

we were given
2 Pi / (Tv) = 5 * 2 Pi / (Tp)

so plug in and solve for Tp

Answer 14

with me so far?

Answer 15

Tp = 437.66 hours

Answer 16

the system says incorrect

Answer 17

does it say to round

Answer 18

no, only the word "incorrect" appears

Answer 19

oh i multiplied by 5, not 5.28

Answer 20

462.165

Answer 21

no, still not right, I'll skip it for now

Answer 22

i did it also the other way the person above me
462.165 hours is the period of the planet

Answer 23

Incorrect. Note that the speedometer only reports the speed of the vehicle relative to the ground. now there is something more

Answer 24

@surry99 please check

Answer 25

145 = 2020 * w_v

145/2020 = w_v

w_p = w_v/5.28

w_p = (145/2020) / 5.28

T = 2pi / (w_p)

Answer 26

T = 462.1651118

Answer 27

Ah yes, forgot about what the speedometer measures ...silly me. hang on.

Answer 28

Is the answer 374 hours

Answer 29

V vehicle = V planet + V vehicle relative to planet
r x omega v = r omega p + 145
and omega vehicle = 5.28 x omega planet so:

r x (5.28 omega p) = r x omega p + 145
solving for omega p...
omega p = 145/(4.28 x r)

we get omega p = 145 km/hr/(4.28 x 2020 km) = .0167/hr
then T = 2 x pi/omega p = 6.28/.0167 /hr = 376 hours

Answer 30

hmm

Answer 31

thankyou surry for your help @surry99

Answer 32

It was a pleasure working with you @perl